c Dr Oksana Shatalov, Spring 2015
1
1
Mathematical Reasoning
Prove: “For every odd integer n, the integer 3n+7 is even.”
1.1
Statements
DEFINITION 1. A statement is any declarative sentence that is either true or false.
A statement cannot be neither true nor false and it cannot be both true and false.
1. The integer 5 is odd.
2. The integer 24 is prime.
3. 15 + 7 = 22
4. Apple manufactures computers.
5. Apple manufactures the world’s best computers.
6. Did you buy IBM?
7. I am telling a lie.
8. What happen when Pinocchio says: “My nose will grow now”?
An open sentence is any declarative sentence containing one or more variables that is not
a statement but becomes a statement when the variables are assigned values.
9. x + 5 = 7
10. He is a student.
11. x2 + y 2 = 1
c Dr Oksana Shatalov, Spring 2015
2
An open sentence can be made into a statement by using quantifiers.
Universal : ∀x means for all/for every assigned value a of x.
Existential : ∃x means that for some assigned vales a of x.
Once a quantifier is applied to a variable, then the variable is called a bound variable. The
variable that is not bound is called a free variable.
12. For every real number x, x + 5 = 7.
Quantifiers:
13. The area of a rectangle is its length times its width.
Quantifiers:
14. A triangle may be equilateral.
Quantifiers:
15. 15 − 5 = 10
Quantifiers:
16. The sum of an even integer and an odd integer is even.
Quantifiers:
17. All positive real numbers have a square root.
Quantifiers:
18. A real-valued function that is continuous at 0 is not necessarily differentiable at 0.
Quantifiers:
c Dr Oksana Shatalov, Spring 2015
3
NEGATIONS
DEFINITION 2. If P is a statement, then the negation of P , written ¬P (read “not P ”),
is the statement “P is false”.
19. All continuous functions are differentiable.
20. P : Barack Obama won the Nobel Peace Prize.
21. P :
53 = 120
22. P (x) :
23. P (x, y) :
¬P :
x2 + x + 1 = 0
¬P (x) :
x4 + y 4 = 0
¬P (x, y) :
24. P : If n is an odd integer then 3n + 7 is odd.
¬P
25. P : Every car on the parking lot #47 was with valid permit.
¬P
26. P :There exist real numbers a and b such that (a + b)2 = a2 + b2 .
¬P
Rules to negate statements with quantifiers:
c Dr Oksana Shatalov, Spring 2015
4
27. P : For every even integer n there exists an integer m such that n = 2m.
¬P
28. P : There exists a positive integer n such that m(n + 5) < 1 for every integer m.
¬P
EXAMPLE 3. P : If n is an integer and n2 is a multiple of 4 then n is a multiple of 4.
Question: Is the following “proof ” valid?
Let n = 6. Then n2 = 62 = 36 and 36 is a multiple of 4, but 6 is not a multiple of 4. Therefore,
the statement P is FALSE.
1.2
Compound Statements
1. P : Some math tests are long.
2. Q: Some math tests are difficult.
Logical connectivity write read
Conjunction
P∧Q P and Q
Disjunction
P∨Q P or Q
meaning
Both P and Q are true
P is true or Q is true
c Dr Oksana Shatalov, Spring 2015
5
TRUTH TABLES
P : Ben is a student.
Q: Ben is a teaching assistant.
P
Q
P ∧Q
Q∧P
P ∨Q
Q∨P
EXAMPLE 4. Rewrite the following open statements using disjunction or conjunction.
(a) P (x) :
|x| ≥ 10.
(b) P (x) :
|x| < 10.
Two compound statements are logically equivalent (write “≡”) if they have the same truth
tables, which means they both are true or both are false.
EXAMPLE 5. Let P and Q be statement forms. Determine whether the compound statements
¬P ∧ Q and ¬P ∨ Q are logically equivalent (i.e. both true or both false).
P
Q
c Dr Oksana Shatalov, Spring 2015
6
Some Fundamental Properties of Logical Equivalence
THEOREM 6. For the statement forms P , Q and R,
• ¬(¬P ) ≡
• Commutative Laws
P ∨Q≡
P ∧Q≡
• Associative Laws
P ∨ (Q ∨ R) ≡
P ∧ (Q ∧ R) ≡
• Distributive Laws
P ∨ (Q ∧ R) ≡
P ∧ (Q ∨ R) ≡
• De Morgan’s Laws
¬(P ∨ Q) ≡ (¬P ) ∧ (¬Q)
¬(P ∧ Q) ≡ (¬P ) ∨ (¬Q)
Proof. Each part of the theorem is verified by means of a truth table.
P
Q
EXAMPLE 7. Rewrite the rules to negate statements with quantifiers in terms of logical equivalence:
¬(∀x, P (x)) ≡
¬(∃x 3 P (x)) ≡
¬(∀x, (P (x) ∨ Q(x)) ≡
¬(∀x, (P (x) ∧ Q(x)) ≡
¬(∃x 3 (P (x) ∨ Q(x)) ≡
¬(∃x 3 (P (x) ∧ Q(x)) ≡
c Dr Oksana Shatalov, Spring 2015
7
EXAMPLE 8. Negate:
P : There exists a prime number p which is greater then 7 and less than 10.
¬P
Tautologies and Contradictions
Tautology: statement that is always true
Contradiction: statement that is always false
¬P
P
T
F
1.3
P ∨ (¬P ) P ∧ (¬P )
Implications
DEFINITION 9. Let P and Q be statements. The implication P ⇒ Q (read “P implies Q”)
is the statement “If P is true, then Q is true.”
EXAMPLE 10. If n is odd, then 3n + 7 is even.
The truth table for implication:
P
T
T
F
F
Q P ⇒Q
T
T
F
F
T
T
F
T
EXAMPLE 11. P : You earn an A on the final exam.
Q: You get an A for your final grade.
P ⇒Q
Different ways of expressing P ⇒ Q :
If P is true, then Q is true
Q is true if P is true
P implies Q
P is true only if Q is true
P is sufficient for Q
Q is necessary for P .
c Dr Oksana Shatalov, Spring 2015
8
EXAMPLE 12. For a triangle T , let
P (T ) : T is equilateral
Q(T ): T is isosceles.
State P (T ) ⇒ Q(T ) in a variety of ways:
EXAMPLE 13. P : The function f (x) = sin x is differentiable everywhere.
Q: The function f (x) = sin x is continuous everywhere.
P ⇒Q
Q⇒P
c Dr Oksana Shatalov, Spring 2015
9
Proving Statements Containing Implications.
Properties of Integers:
FACT 1 The negative of every integer is an integer.
FACT 2 The sum (and difference) of every two integers is an integer.
FACT 3 The product of every two integers is an integer.
FACT 4 Every integer is either even, or odd.
• DIRECT PROOF
– Assume that P is true.
– Draw out consequences of P .
– Use these consequences to prove Q is true.
EXAMPLE 14. If n is an even integer, then 5n5 is an even integer.
EXAMPLE 15. Evaluate the proposed proof of the following result:
If a is an even integer and b is an odd integer, then 3a − 5b is odd.
Proof. Let a be an even integer and b be an odd integer. Then a = 2n and b = 2n + 1 for
some integer n. Therefore,
3a − 5b = 3(2n) − 5(2n + 1) = 6n − 10n − 5 = −4n − 5 = 2(−2n − 2) − 1.
Since −2n − 2 is an integer, 3a − 5b is odd. c Dr Oksana Shatalov, Spring 2015
EXAMPLE 16. The sum of every two odd integers is even.
EXAMPLE 17. Let x be an integer. If 5x − 7 is odd, then 9x + 2 is even.
10
c Dr Oksana Shatalov, Spring 2015
• PROOF BY CASES
EXAMPLE 18. If n is an integer, then n2 + 3n + 4 is an even integer.
Hint: Use the following fact: “Every integer number is either even. or odd.”
11
c Dr Oksana Shatalov, Spring 2015
Negating an implication:
12
Counterexamples
¬(P ⇒ Q) ≡ P ∧ (¬Q)
REMARK 19. The negation of an implication is not an implication!
EXAMPLE 20. Negate the statement: “For all x, P (x) ⇒ Q(x).”
The value assigned to the variable x that makes P (x) true and Q(x) false is called a counterexample to the statement “For all x, P (x) ⇒ Q(x).”
EXAMPLE 21. Disprove the following statement:
If a real-valued function is continuous at some point, then this function is differentiable there.
c Dr Oksana Shatalov, Spring 2015
13
Necessary and Sufficient Conditions
P ⇒ Q can be expressed as
P is sufficient for Q.
or
Q is necessary for P .
Equivalently,
In order for Q to be true it is sufficient that P be true.
or
Q must be true in order to P to be true.
EXAMPLE 22. Consider the following open sentences
P (x) : x is a multiple of 4.
Q(x) : x is even. Complete:
• “ ∀x, P (x) ⇒ Q(x)” is
.
• P (x) is a
condition for Q to be true.
• Q(x) is a
condition for P (x) to be true.
• Q(x) is not a
condition for P (x) to be true.
EXAMPLE 23. Consider the following open sentences
P (f ) : f is a differentiable function.
Q(f ) : f is a continuous function.
Complete:
• “ ∀f, P (f ) ⇒ Q(f )” is
.
• “ ∀f, Q(f ) ⇒ P (f )” is
.
• Q(f ) is a
condition.
• P (f ) is a
condition for f to be differentiable, but not a
condition for f to be continuous.
REMARK 24. Note however, if P ⇒ Q is true, then it is not necessary that P is true in order
for Q to be true. Even if Q is true, P may be false.
c Dr Oksana Shatalov, Spring 2015
1.4
14
Contrapositive and Converse
Contrapositive
DEFINITION 25. The statement ¬Q ⇒ ¬P is called the contrapositive of the statement
P ⇒ Q.
EXAMPLE 26. Let P and Q be statement forms. Prove that ¬Q ⇒ ¬P is logically equivalent
to P ⇒ Q.
P
Q P ⇒Q
¬Q ¬P
¬Q ⇒ ¬P
Methods to prove an implication P ⇒ Q (continued)
• CONTRAPOSITIVE PROOF (based on the equivalence (P ⇒ Q) ≡ (¬Q ⇒ ¬P ))
– Assume that ¬Q is true.
– Draw out consequences of ¬Q.
– Use these consequences to prove ¬P is true.
– It follows that P ⇒ Q.
REMARK 27. If you use a contrapositive method, you must declare it in the beginning
and then state what is sufficient to proof.
EXAMPLE 28. Let x be an integer. If 5x − 7 is even, then x is odd.
c Dr Oksana Shatalov, Spring 2015
15
Converse
DEFINITION 29. The statement Q ⇒ P is called a converse of the statement P ⇒ Q.
Question: Are the statements P ⇒ Q and Q ⇒ P logically equivalent?
EXAMPLE 30. If m and n are odd integers then m + n is even.
Biconditional “⇔”
DEFINITION 31. The statement P ⇔ Q (or P iff Q) is the statement (P ⇒ Q) ∧ (Q ⇒ P ).
P
T
T
F
F
Q
T
F
T
F
P ⇒Q Q⇒P
P ⇔Q
THEOREM 32. Let ~a and ~b be two non zero vectors. Then ~a is orthogonal to ~b iff ~a · ~b = 0.
c Dr Oksana Shatalov, Spring 2015
THEOREM 33. Let n be an integer. Then n is even if and only if n2 is even.
Proof.
REMARK 34. (P ⇔ Q) ≡ (¬P ⇔ ¬Q)
COROLLARY 35. Let n be an integer. Then n is odd iff n2 is odd.
16
c Dr Oksana Shatalov, Spring 2015
17
Methods to prove an implication P ⇒ Q (continued)
EXAMPLE 36. Let S and C be statement forms. Prove that ¬S ⇒ (C ∧ ¬C) is logically
equivalent to S.
• PROOF BY CONTRADICTION
– Assume that P is true.
– To derive a contradiction, assume that ¬Q is true.
– Prove a false statement C, using negation ¬(P ⇒ Q) ≡ (P ∧ ¬Q).
– Prove ¬C. It follows that Q is true. (The statement C ∧ ¬C must be false, i.e. a
contradiction.)
PROPOSITION 37. If m and n are integers, then m2 − 4n 6= 2.
Proof.
c Dr Oksana Shatalov, Spring 2015
18
PROPOSITION 38. Let be a, b, and c be integers. If a2 + b2 = c2 then a or b is an even integer.
Proof.
c Dr Oksana Shatalov, Spring 2015
19
DEFINITION 39. A real number x is rational if x =
Also, x is irrational if it is not rational, that is
PROPOSITION 40. The number
√
2 is irrational.
m
n
for some integer numbers m and n.